Chapter 7.5 - Eye and Vision

In the previous section we saw lens formula and magnification. In this section, we will learn about the Eye and vision.
The structure of the eye is shown in the fig. below. It is obtained from wikimedia commons.

Let us see the important features:
1. The amount of light reaching the lens is controlled by the iris. 
• When the surrounding light is less, the iris expands to let in more light which may be required for clear vision
• When the surrounding light is more, the iris contracts to let in only the just required light for clear vision
2. The lens in the front of the eye is a convex lens
• So light rays emerging from an object will have to pass through this convex lens
• When they pass through the lens, they will converge. This is shown schematically in the fig.7.28 below:

Fig.7.28

[Note that, in the fig.7.28, two light rays emerging from a 'point object' is shown. This is for convenience only. For larger objects, more rays will have to be drawn]
3. In fig.7.28, we can see that the image is formed exactly on the rear side of the eyeball. 
• At this rear side, the retina is present. This is shown in yellow colour in the previous fig.7.27 above.
• So we can say, the image is formed exactly on the retina. 
• Optical nerves of the retina send the sense of vision to the brain. Due to the activity of brain, we see the object
4. We now know how we see 'O'.
■ But what if the position of 'O' change?
• The lens cannot change it's position
• The retina cannot change it's position
• So the only option is to 'change the focal length' of the lens
    ♦ 'Change in focal length' is achieved by 'changing the curvature' of the lens
    ♦ 'Change in curvature' is achieved by the expansion and contraction of the ciliary muscles
• Thus we can say:
The eye has the ability to form an image, whatever be the position of the object

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The ability of the eye to form an image on the retina by adjusting the focal length of the lens in the eye, by varying the curvature of the lens whatever be the position of the object, is the power of accommodation

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• If we hold a book very close to our eyes, we will not be able to read the letters. 
• That means, in spite of the 'power of accommodation', there are some limitations. 
• If the letters are too close, a clear image (as shown in fig.7.28) cannot be formed. 
• The lens cannot accommodate such an extreme condition
■ The nearest point at which one can see an object clearly is called the Near point. The near point of a healthy eye is 25 cm
• In the same way, in the case of 'very far away objects' also, there are limitations

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Defects in vision

Case 1:
Consider fig.7.29 below. We will do an analysis of the fig.:

Fig.7.29

1. In fig.7.29(a), after passing through the lens, the rays are indeed converging. 
• But the point of convergence is not on the retina
• The point of convergence is behind the retina. What could be the reason for this?
2. It may be due to any one or both of the two reasons given below:
(i) The power of the lens decreased
• 'Decrease in the power' of lens means that the lens is not able to 'cause enough bending' of light rays
• In the fig.7.29(a) we see that, if the lens is able to cause a little more bending, the light rays would surely intersect at the retina  
(ii) The eyeball contracted so that the retina moved forward
3. What will be the result when such defects occur?
• We see that a perfect image will be formed behind the retina. 
• But that image has no use. The retina can send signals based on only those images which fall exactly on it
■ We see that rays 'nearly' converge on the retina. But the convergence is not perfect. 
• So the image formed on the retina will not be perfect. It will be 'blurred'. 
• So a person with such defects will see only a blurred image
4. What is the remedy for this defect?
Consider fig.7.29(b). After passing through the lens, two sets of rays are shown
(i) A set consisting of two white rays
    ♦ This is the same shown in fig.a 
(ii) A set consisting of two green rays
• If we follow the white set, we can see that the problem will not be remedied. 
    ♦ They intersect behind the retina
• If we follow the green set, we can see that the problem will be remedied. 
    ♦ They intersect exactly on the retina
5. So we want the green set. That is., after passing through the lens, the rays must follow the green lines. How can we make that happen?
• We notice that, the green rays 'converge more' than the white rays. That is why they are able to intersect exactly on the retina
• So if we can make the rays to 'converge more', the defect will be remedied
6. How can we make them to 'converge more'?
Ans: By using a converging lens. 
A convex lens is a converging lens. So we can use it. It is shown in fig.7.29(c)
7. So we must put a convex lens in front of the eye to remedy this defect
• The convex lens in front of the eye, will create some degree of convergence even before the light enters the eye
• Then the lens in the eye will need to do only a 'lesser work' for making the rays intersect exactly on the retina
• The power of the 'convex lens to be used' will be prescribed by the doctor after examination

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Case 2:
Consider fig.7.30 below. We will do an analysis of the fig.:

Fig.7.30

1. In fig.7.30(a), after passing through the lens, the rays are indeed converging. 
• But the point of convergence is not on the retina
• The point of convergence is in front of the retina. What could be the reason for this?
2. It may be due to any one or both of the two reasons given below:
(i) The lens has increased power
• 'Increase in the power' of lens means that the lens 'cause a bending, which is more than what is required'.
(ii) The eyeball is longer than normal so that the retina moved backwards
3. What will be the result when such defects occur?
• We see that a perfect image will be formed in front of the retina. 
• But that image has no use. The retina can send signals based on only those images which fall exactly on it
■ We see that rays 'nearly' converge on the retina. But the convergence is not perfect. 
• So the image formed on the retina will not be perfect. It will be 'blurred'. 
• So a person with such defects will see only a blurred image
4. What is the remedy for this defect?
Consider fig.7.30(b). After passing through the lens, two sets of rays are shown
(i) A set consisting of two white rays
    ♦ This is the same shown in fig.a 
(ii) A set consisting of two green rays
• If we follow the white set, we can see that the problem will not be remedied. 
    ♦ They intersect in front of the retina
• If we follow the green set, we can see that the problem will be remedied. 
    ♦ They intersect exactly on the retina
5. So we want the green set. That is., after passing through the lens, the rays must follow the green lines. How can we make that happen?
• We notice that, the green rays 'converge less' than the white rays. That is why they are able to intersect exactly on the retina
• So if we can make the rays to 'converge less', the defect will be remedied
6. How can we make them to 'converge less'?
Ans: 'Converging less' is equivalent to 'diverging more'
• So, by using a diverging lens, we can remedy this defect. 
• A concave lens is a diverging lens. So we can use it
7. So we must put a concave lens in front of the eye to remedy this defect
• The concave lens in front of the eye, will create some degree of divergence even before the light enters the eye
• Then the rays will intersect exactly on the retina, even if the lens of the eye has increased power
• The power of the 'concave lens to be used' will be prescribed by the doctor after examination

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The two defects that we saw above are given special names:
Case 1:
This is called Hypermetropia or Far-sightedness
The definition is:
■ Since the image is formed behind the retina, instead of being formed at the retina, even though distant objects are clearly seen, nearer objects cannot be seen. This defect of the eye is called Hypermetropia or Far-sightedness
• In the above definition, we see a peculiarity:
• It is said that, 'distant objects can be clearly seen'. That means people with hypermetropia:
    ♦ Can see distant objects
    ♦ Cannot see near objects 
• How is that possible?
We will write the answer in steps:
1. Consider the fig.7.31(a) below:

Fig.7.31

• It shows an eye with out any defects. 
• The white rays from the near object O are neatly converging on the retina
2. Another set of green rays are shown in the same fig.a 
• Those rays are shown to be parallel to each other. 
3. Why are they parallel to each other? 
• The answer is that, those rays come from distant objects, and such rays will be always parallel to each other
4. In fig.7.31(b), two additional arcs are shown
• One magenta arc and one yellow arc
• The magenta arc shows the angle between incident white ray and refracted white ray
• The yellow arc shows the angle between incident green ray and refracted white ray
• Obviously, the angle indicated by the magenta arc is larger than that indicated by the yellow arc
5. So now compare the incident green set and the incident white set in fig.b 
• We can see that, 'bending through a greater angle' is required to bring the white rays together on to the retina
• In other words, when compared to the white rays, a 'lesser effort' by the lens is sufficient to bring the green rays together on to the retina
6. Since 'lesser effort' is sufficient, even eyes with hypermetropia can make the rays from distant objects to intersect on the retina
• In other words, persons with hypermetropia will be able to see distant objects with out external help. 
• While they cannot see near objects clearly. So another name for this defect is Far-sightedness.

Case 2:
This is called Myopia or Near-sightedness
The definition is:
■ Since the image is formed in front of the retina, instead of being formed at the retina, even though nearby objects are clearly seen, distant objects cannot be seen. This defect of the eye is called Myopia or Near-sightedness
• In the above definition, we see a peculiarity:
It is said that, 'near objects can be clearly seen'. That means people with myopia:
    ♦ Can see near objects
    ♦ Cannot see distant objects  
• How is that possible?
The answer is just opposite to the one that we saw for case 1. But we will write the steps again:
1. Consider the same fig.7.31 above.
• It shows an eye with out any defects. 
• The white rays from the near object O are neatly converging on the retina
2. Another set of green rays are shown in the same fig.a 
• Those rays are shown to be parallel to each other. 
3. Why are they parallel to each other?
• The answer is that, those rays come from distant objects, and such rays will be always parallel to each other
4. In fig.7.31(b), two additional arcs are shown
• One magenta arc and one yellow arc
• The magenta arc shows the angle between incident white ray and refracted white ray
• The yellow arc shows the angle between incident green ray and refracted white ray
• Obviously, the angle indicated by the magenta arc is larger than that indicated by the yellow arc
5. So now compare the green set and the white set. 
• We can see that, 'more bending' is required to bring the white rays together on to the retina
• In other words, when compared to the white rays, a 'lesser effort' by the lens is sufficient to bring the green rays together on to the retina
6. Since 'lesser effort' is sufficient, the eyes with myopia (which has increased converging power) will be converging the green rays before they reach the retina. 
• Converging of rays before reaching the retina is not useful
7. The white rays require 'more bending'. So the extra converging power available will be compensated. 
• That is., the white rays will converge on the retina. Which means that near objects can be clearly seen without external help. So another name for this defect is Near-sightedness.

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Astigmatism

1. Normally, the lens of the eye is smoothly curved in all directions. 
2. But if the curvature is not smooth, the surface of the lens will be irregular. 
3. In such a situation, the light rays from both near and far objects will not be refracted properly. 
4. They will not intersect on the retina. So both near and far objects will appear to be blurred. 
5. This condition is called astigmatism. 
6. This can be remedied by using cylindrical lens of appropriate power.

Presbyopia

1. We have seen that, the rays from near objects need 'greater bending'.
2. For this, the power of the lens of the eye will have to be increased. 
• This is achieved with the help of the ciliary muscles
3. But in elderly people, the ciliary muscles may not be able to provide the necessary curvature to the lens.
4. So elderly people will have to hold the objects at a distance greater than 25 cm, to see them properly. This condition is called presbyopia. 
5. This can be remedied by using convex lens of appropriate power

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Doctor's prescription
• The prescription made by the doctor will show the power of the required lens for rectifying the defects of the eye.
• Power of a lens is related to it's focal length
• Let us see an example:
1. Focal length of a lens is 20 cm
2. Express this focal length in metres:
• We have: 20 cm = 0.20 m
3. Take the reciprocal
• We have: Reciprocal of 0.20 = ¹⁄_0.02 = ¹⁰⁰⁄₂₀ = 5
4. This '5' is the power of the lens.
We can write the definition of power:
■ Power of a lens is the reciprocal of focal length expressed in metres. 
• Power = ¹⁄_f
    ♦ Where f is the focal length expressed in metres
• Unit of power is dioptre. It is represented by D
■ Power of convex lens is positive and that of a concave lens is negative

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But why do we take the reciprocal of the focal length?
The answer can be written in steps:
1. Consider fig.7.13 that we saw in a previous section. For convenience, it is shown again below:

2. Consider the convex lens in fig.a. It is clear that convex lens is used for converging light rays
• If the convex lens has greater ability to converge rays, we can say that it has greater power
• If the convex lens has lesser ability to converge rays, we can say that it has lesser power
3. But from the fig.a, it is clear that:
• If greater convergence is achieved, the focus F will move towards the lens
    ♦ That is., focal length will decrease 
• If lesser convergence is achieved, the focus F will move away from the lens
    ♦ That is., focal length will increase
4. So we can conclude. In the case of a convex lens:
• Greater power means lesser focal length
• Lesser power means greater focal length
5. So there is an 'inverse relation'
• Such a relation can be expressed only if we put 'f' in the denominator.
• Because in the relation Power = ¹⁄_f, 
    ♦ When f increases, power decreases  
    ♦ When f decreases, power increases
■ So we indeed have to take the reciprocal of 'f'

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6. Now consider the concave lens in fig.b. It is clear that concave lens is used for diverging light rays
• If the concave lens has greater ability to diverge rays, we can say that it has greater power
• If the concave lens has lesser ability to diverge rays, we can say that it has lesser power
7. But from the fig.b, it is clear that:
• If greater divergence is achieved, the focus F will move towards the lens
    ♦ That is., focal length will decrease 
• If lesser divergence is achieved, the focus F will move away from the lens
    ♦ That is., focal length will increase
8. So we can conclude. In the case of a concave lens:
• Greater power means lesser focal length
• Lesser power means greater focal length
9. So there is an 'inverse relation'
• Such a relation can be expressed only if we put 'f' in the denominator. 
• Because in the relation Power = ¹⁄_f, 
    ♦ When f increases, power decreases  
    ♦ When f decreases, power increases
■ So for the concave lens also, we indeed have to take the reciprocal of 'f'

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Solved example 7.6
The powers of the lenses in the prescription of the doctor are +1.50 D and +1 D. What are the focal lengths of these two lenses? What type of lenses are they?
Solution:
Case 1:
1. We have: Power = +1.5 D = ¹⁄_f
    ♦ Where f is the focal length expressed in metres 
2. So f = ¹⁄_+1.5 = ⁺¹⁰⁄₁₅ = ⁺²⁄₃ = +0.666 metres = +66.6 cm
3. Since the focal length is positive, it is a convex lens
Case 2:
1. We have: Power = +1 D = ¹⁄_f
    ♦ Where f is the focal length expressed in metres 
2. So f = ¹⁄₊₁ = +1 metre = +100 cm
3. Since the focal length is positive, it is a convex lens

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In the next chapter, we will see Current  electricity.

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Chapter 7.4 - The Lens Formula and Magnification

In the previous section we saw that the ray diagrams when the object is placed at different positions in front of the lens. In this section, we will see an experiment based on those ray diagrams.

Before doing the experiment, we must do an analysis of the fig.7.26 below.

We will write the analysis in steps:
1. In fig.7.26(a), the object is placed at 6.25 cm to the left of P. So u = -6.25 cm
2. The two rays intersect at 11.11 cm to the right of P. So v = 11.11 cm
3. If we place a screen at this 11.11 cm, we will get a clear image of the object
4. Now consider fig.b
• The lens is kept at the same position.
• The screen is kept at the same position
• But the object is moved. 
5. The image disappears from the screen. Why does it disappear?
• The reason is that, the two rays no longer intersect at 11.11 cm.
• When the object was moved, the point of intersection of the two rays also moved
• So we will not get an image at 11.11 cm
6. Why does the point of intersection move?
The answer can be written in steps: 
(i) The top most ray parallel to the principal axis is the same in both the figs
(ii) But consider the other ray passing through P:
    ♦ In fig.a, it passes through (-6.25, 2.25) and P(0,0) 
    ♦ In fig.b, it passes through (-7.25, 2.25) and P(0,0)
    ♦ [The y coordinate 2.25 is the height of the object]
• So this ray is not the same in figs (a) and (b)
• Thus the point of intersection will differ
■ Now, to make the image appear again, we will have to move the screen
7. Based on the above analysis, we can write:
■ If the object is moved keeping the lens at the same position, the screen will also have to be moved to get the image

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Now we can begin the experiment:
1. Put a lighted candle in front of a convex lens
2. Put a screen on the other side of the lens
• Adjust the position of the screen to get a clear image of the candle
3. Note down u and v
• They are recorded as -30 and 60 respectively in the first row of the table below:

4. With out changing the position of the lens, move the candle.
• Let it be moved by 10 cm to towards the left. Then new u = -40 cm
5. Adjust the position of the screen to obtain a clear image of the candle
• New v is obtained as 40 cm. These are recorded in the second row
6. Repeat the trial. 
• u changes to -50 and v changes to 33.33
7. The observations are over. Now we do some calculations:
• Fill up the fourth column with '¹⁄_u' values
• Fill up the fifth column with '¹⁄_v' values
8. Subtract '¹⁄_u' from '¹⁄_v' and fill up the sixth column
• We see that the heading of this sixth column is '[¹⁄_f = ¹⁄_v - ¹⁄_u]'
• It is an equation. According to this equation:
• When we Subtract '¹⁄_u' from '¹⁄_v', we get the reciprocal of the focal length 'f'
9. So from the first trial we get: Reciprocal of 'f' is 0.05
    ♦ That is., ¹⁄_f = 0.05 ⟹ f = ¹⁄_0.05 = 20  
• From the second trial we get: Reciprocal of 'f' is 0.05
    ♦ That is., ¹⁄_f = 0.05 ⟹ f = ¹⁄_0.05 = 20
• From the third trial we get: Reciprocal of 'f' is 0.050003
    ♦ That is., ¹⁄_f = 0.050003 ⟹ f = ¹⁄_0.050003 = 19.9988
10. Note that, 0.05 and 0.050003 are nearly equal
• Their reciprocals will also be nearly equal
• We find that '20' and '19.9988' are indeed nearly equal     
11. The equation '[¹⁄_f = ¹⁄_v - ¹⁄_u]' can be easily simplified to the form: '[f = ^uv⁄_(u-v) ]'
• Using this, we can directly calculate 'f'. The fourth, fifth and sixth columns can be avoided. 
• The results obtained in this way are recorded in the seventh column
• Note that, we get the same values for 'f'
12. Which ever be the method, we must calculate the average of the results from all the trials;
• Trial 1 gives: f = 20
• Trial 2 gives: f = 20
• Trial 3 gives: f = 19.9988
■ So average f = ^{(20+20+19.9988)}⁄₃ = 19.9996 ≈ 20 cm

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■ The equation '[¹⁄_f = ¹⁄_v - ¹⁄_u]' that we saw above is called the lens equation.
It's derivation can be seen here.
• Simplifying it, we got: [f = ^uv⁄_(u-v) ]
• Two more forms can be obtained:
[v = ^uf⁄_(u+f)] and [u = ^fv⁄_(f-v)]
• These equations can be used for solving problems

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Solved example 7.1
When an object is placed at a distance of 15 cm from a convex lens, a real image is formed at a distance of 30 cm. What is the focal length of the lens?
Solution:
1. Given that, the object and image is real. 
• For a convex lens, object and it's real image will always be on opposite sides 
• So one distance will be negative and the other, positive. 
2. We will assume that object is on the left side and image is on the right side of the lens. Thus we can write:
u = -15 cm, v = 30 cm
3. We have: f = ^uv⁄_(u-v)
Substituting the values, we get: f = ^(-15×30)⁄_(-15-30) = ^-450⁄_-45 = 10 cm
4. Note that we got a positive value for 'f'
• The focal length of convex lens is always positive
• The focal length of concave lens is always negative

Solved example 7.2 
The focal length of a concave lens is 20 cm. If an object is kept at a distance of 30 cm from the lens, find out the distance of the image formed
Solution:
1. For a concave lens, both object and image will always be on the same side of the lens. We will assume that, they are both on the left side of the lens. 
• Then we can write: u = -30
2. The focus of a concave lens is virtual. So it's focal length is negative. 
• Then we can write: f = -20 cm
3. We have: v = ^uf⁄_(u+f)
• Substituting the values, we get: v = ^(-30×-20)⁄_(-30-20) = ⁺⁶⁰⁰⁄_-50 = -12 cm
4. Note that a negative value is obtained for v
• u was also given a negative value
• So both the object and image are indeed on the same side (left) of the lens

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Magnification

• When we discussed ray diagrams, we saw the following:
(i) The image may be enlarged than the object
(ii) The image may be of the same size as the object
(iii) The image may be diminished than the object
• Consider the case 1. We would want to know 'how much enlargement' occurred?
We will write the steps:
1. Both height and width will be enlarged by the same factor. Let 'm' be the factor
2. Then we can say: [Height of object] × m = [Height of image]
• Rearranging this equation, we get: m = ^{Height of image}⁄_{Height of object}
3. This gives us an easy method to find 'm'
• It is clear that, 
(i) If 'm' is greater than 1, the image will be enlarged 
(ii) If 'm' is equal to 1, the image will be of the same size
(iii) If 'm' is less than 1, the image will be diminished
4. This 'm' is called magnification. 
■ We can write the definition:
Magnification is the ratio of the height of the image to the height of the object. It shows how many times the image is as large as the object
5. Magnification = ^{Height of image}⁄_{Height of object} = ^h_i⁄_ho.
• So to find 'm':
    ♦ We need to measure the height of the object (h_o)
    ♦ We need to measure the height of the image (h_i)
• However, there is another method to calculate 'm'
• This is based on the fact that (^h_i⁄_ho) will be equal to (^v⁄_u)
• So we can write: m = ^h_i ⁄_ho = ^v⁄_u.
• Obviously it is easier to measure u and v. So 'm' can be calculated easily 
• The mathematical steps which prove that '(^h_i ⁄_ho) is equal to (^v ⁄_u)' can be seen here.

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Solved example 7.3
When an object of height 3 cm is placed at a distance of 30 cm from a lens, a real image is formed at a distance of 60 cm. Find out the height of the image
Solution:
1. Given that, the image is real. So the object and image are on the opposite sides of the lens
• Let the object be on the left side. Then u = -30 cm
• Let the image be on the right side. Then v = 60 cm
2. We have: magnification = ^v⁄_u = ⁶⁰⁄_-30 = -2
3. We also have: magnification = ^h_i ⁄_ho = ^h_i ⁄₃ =  ⁶⁰⁄_-30 = -2
4. Equating the results in (2) and (3), we get: ^h_i ⁄₃ = -2
⟹ h_i  = -6 cm 
• The negative sign indicates that the height of the image is measured below the principal axis. In other words, the image is formed below the principal axis

Solved example 7.4
A concave lens of focal length 10 cm produces an image which is half the size of the object. How far away is the object from the lens? Find also the position and nature of the image.
Solution:
1. Focal length of a concave lens should be taken as negative. So we can write: f = -10 cm
2. Given that image is half the size of the object. So we can write: h_i = h_o × ¹⁄₂
• So m = ^h_i ⁄_ho = ¹⁄₂
3. We also have: magnification = ^v⁄_u
4. Equating the results in (2) and (3), we get: ^v⁄_u = ¹⁄₂
⟹ u = 2v
5. We have: ¹⁄_f = ¹⁄_v - ¹⁄_u
Substituting u = 2v and f = -10, we get:
¹⁄_-10 = ¹⁄_v - ¹⁄_2v 
⟹ ¹⁄_-10 = ^(2-1)⁄_2v = ¹⁄_2v =
⟹ 2v = -10 ⟹ v = -5 cm
6. Then u = 2v = -10 cm
7. Both u and v are negative. So both are on the left side of the lens
• The object is 10 cm from optic centre P
• The image is 5 cm from optic centre P   
• For a concave lens, image will always be virtual

Solved example 7.5
An object of height 3 cm is placed in front of a convex lens of focal length 10 cm at a distance of 15 cm.
(a) What is the distance of the image formed?
(b) What is the nature of the image?
(c) What is the height of the image?
Solution:
Part (a): We have to find v
We have: v = ^uf⁄_(u+f)
• Substituting the values, we get: v = ^(-15×10)⁄_(-15+10) = ^-150⁄_-5 = 30 cm
Part (b): 
1. We obtained a positive value for v. The distance u was taken as negative 
• That means, image is formed on the other side of the lens
• The image formed by a convex lens in this manner will be always real and inverted
Part (c):
• We have: magnification m = ^h_i ⁄_ho = ^v⁄_u = ³⁰⁄_-15 = -2
⟹ h_i = -2 × h_o = 2 × 3 = -6 cm
• The negative height shows that the image is inverted 

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In the next section, we will see Eye and vision.

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Chapter 7.3 - Image formation and Ray diagrams

In the previous section we saw that the rays emerging from the object, passes through the lens and then converge to give points on the image. To draw the exact paths taken by the rays, we must be familiar with some rules. They are given below:
Rule 1:
• When a light ray passes through the optic centre of a thin lens, it does not undergo any deviation. This is shown in figs.7.16 (a) and (b) below:

Fig.7.16

• We can see that the rays are passing through P. The lens is not capable of causing any deviation to such rays
Rule 2:
A ray of light falling parallel to the principal axis of a convex lens passes through the principal focus after refraction. This is shown in fig.7.17 below:

Fig.7.17

Rule 3:
A ray of light falling parallel to the principal axis of a concave lens appears to emerge from the focus on the same side of the lens. This is shown in fig.7.18 below:

Fig.7.18

• Note that, after refraction, the ray diverges away. We can only trace it's path backwards by drawing a dashed line on paper. When we draw such a dashed line, we will reach the focus F 
Rule 4:
A ray of light passing through the principal focus of a convex lens passes parallel to the principal axis after refraction. This is shown in fig.7.19 below:

Fig.7.19

This can be explained using the following steps:
1. Light rays emerge in all directions from an object
2. Among those rays in (1) abive, some will pass through F
3. Among those rays in (2) above, some will fall on the lens
4. All rays in (3) will undergo deviation. 
• After the deviation, they will become parallel to the principal axis
• This can be considered as the reverse of Rule 2
Rule 5:
• All the light rays coming from an 'object at infinity' (or a 'distant object') are considered to be parallel to each other. 
• We have seen that such parallel rays will converge at F. See fig.7.13 of the previous section

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Now we are ready to draw ray diagrams. The ray diagrams will help us to predict three items:
(i) Position of the image
(ii) Nature of the image
(iii) Size of the image

Case 1: Object beyond 2F
This is shown in fig.7.20 below:

Fig.7.20

It is convenient to draw the ray diagram on a graph paper. Let us write the steps:
1. The optic centre P must be placed at the 'origin of the graph'
• 'Origin of the graph' is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm. 
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 9.5 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the object from P is denoted as 'u'
■ So we can write: u = -9.5 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
    ♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
    ♦ After refraction, it will pass through F
(ii) The second one passes through P
    ♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
    ♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 6.9 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = +6.9 cm or simply 7 cm 
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is +2.25 cm or simply 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 1.63 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -1.63 cm
10. Considering the numeric values only, we have:
• Height of the object = 2.25 cm
• Height of the image = 1.63 cm
• So the height of the image has decreased.
• Not only the height, the thickness also will decrease. So a thinner yellow line and a smaller square are shown as the image
11. We can write the conclusion:
■ When the object is placed beyond 2F:
(i) Position of the image: Between F and 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Diminished

Case 2: Object at 2F
This is shown in fig.7.21 below:

As before, we will draw the ray diagram on a graph paper. 
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm. 
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 2F
■ So we can write: u = -8 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
    ♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
    ♦ After refraction, it will pass through F
(ii) The second one passes through P
    ♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
    ♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 8 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = 8 cm
• Note that this is the distance of 2F (on the other side) from P
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is also 2.25 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -2.25 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = -2.25 cm
• So the height of the image has not changed.
• Not only the height, the thickness also will not change.
11. We can write the conclusion:
■ When the object is placed at 2F:
(i) Position of the image: At 2F on the other side
(ii) Nature of the image: Real and inverted

(iii) Size of the image: Same size

Case 3: Object between F and 2F
This is shown in fig.7.22 below:

Fig.7.22

As before, we will draw the ray diagram on a graph paper. 
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm. 
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 6.25 units to the left of P
■ So we can write: u = -6.25 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
    ♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
    ♦ After refraction, it will pass through F
(ii) The second one passes through P
    ♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
    ♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 11.11 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = 11.11 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 4 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -4 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 4 cm
• So the height of the image has increased.
• Not only the height, the thickness also will increase. So a thicker yellow line and a bigger square are shown as the image
11. We can write the conclusion:
■ When the object is placed between F and 2F:
(i) Position of the image: Beyond 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Enlarged

Case 4: Object at F
This is shown in fig.7.23 below:

Fig.7.23

As before, we will draw the ray diagram on a graph paper. 
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm. 
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at F. That is., 4 units to the left of P
■ So we can write: u = -4 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
    ♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
    ♦ After refraction, it will pass through F
(ii) The second one passes through P
    ♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
    ♦ It will not undergo any deviation
7. After the refraction, the two rays become parallel to each other
• They will never meet
• Since they are parallel, they will not meet even if we draw them backwards
8. We can write the conclusion:
■ When the object is placed at F:
• Image is not formed

Case 4: Object between F and the lens
This is shown in fig.7.24 below:

Fig.7.24

As before, we will draw the ray diagram on a graph paper. 
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm. 
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 2.5 units to the left of P
■ So we can write: u = -2.5 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
    ♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
    ♦ After refraction, it will pass through F
(ii) The second one passes through P
    ♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
    ♦ It will not undergo any deviation
7. The two rays diverge from each other. They will not meet on the right side of the lens
• But if they are drawn backwards, they will meet at a point on the same side
    ♦ This is shown by the dashed lines
• This meeting point is the image of the 'top of the object'
• From there we can draw the image downwards
8. The position of the image:
The image is formed at 6.66 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = -6.66 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 6 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the image in our present case is 6.0 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 6.0 cm
• So the height of the image has increased.
• Not only the height, the thickness also will increase. So a thicker yellow line and a bigger square are shown as the image
11. We can write the conclusion:
■ When the object is placed between F and lens:
(i) Position of the image: Same side of the object
(ii) Nature of the image: Virtual and erect
(iii) Size of the image: Enlarged

Case 5: Image formation by a concave lens
This is shown in fig.7.25 below:

Fig.7.25

As before, we will draw the ray diagram on a graph paper. 
• The steps are as follows:
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm. 
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. We will not be needing 2F for this case
5. The object is placed at 6.25 units to the left of P
■ So we can write: u = -6.25 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
    ♦ We know what will happen to such a ray. We know it from Rule 3 that we saw above.
    ♦ After refraction, it will diverge away
(ii) The second one passes through P
    ♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
    ♦ It will not undergo any deviation
7. The two rays diverge from each other. They will not meet on the right side of the lens
• But if they are drawn backwards, they will meet at a point on the same side
    ♦ This is shown by the dashed line for ray (i)
    ♦ For ray (ii), the dashed line will fall on that ray itself. So it is not visible
• This meeting point is the image of the 'top of the object'
• From there we can draw the image downwards
8. The position of the image:
The image is formed at 2.21 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = -2.21 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 0.79 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the image in our present case is 0.79 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 0.79 cm
• So the height of the image has decreased.
• Not only the height, the thickness also will decrease. So a thinner yellow line and a smaller square are shown as the image
11. We can write the conclusion:
■ When the object is placed anywhere in front of a concave lens:
(i) Position of the image: Same side of the object between F and the lens
(ii) Nature of the image: Virtual and erect
(iii) Size of the image: Diminished

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In the next section, we will see an experiment based on the above results.

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